Calculate the work done when a force of 20N stretches a spring by 50mm.

Analyze the question: A force of 20 N stretches a spring by 50 mm. We need the work done in stretching it. This is a spring (Hooke's law) problem, so the force is not constant. It rises from 0 to 20 N as the spring stretches.

Concept: Hooke's law gives F = ke, so the work done in stretching a spring is the area under the force-extension graph:

W = ½Fe = ½ke²

Solution:

1. Convert units: e = 50 mm = 0.05 m
2. Force: F = 20 N
3. Work done:

W = ½ × F × e W = ½ × 20 × 0.05 W = 0.5 J

(Check via the spring constant: k = F/e = 20/0.05 = 400 N/m, so W = ½ × 400 × (0.05)² = ½ × 400 × 0.0025 = 0.5 J ✓)

Answer: The work done is 0.5 J, which is option A.

Common mistakes:

• Using W = F × e = 20 × 0.05 = 1.0 J. This is wrong because the force is not constant, and it gives a value that isn't among the options.
• Forgetting to convert mm to m. Using 50 directly would give a huge, wrong value.

Exam tip: Whenever you see a spring being stretched from its natural length, think "half": W = ½Fe. Always convert to metres before you calculate.